What is the integral of 1/((x+3)sqrt((x+3)^2-36)) ?

The integral of 1/((x+3)sqrt((x+3)^2-36)) is 1/6 arcsec(1/6 (x+3))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((x+3)sqrt((x+3)^2-36))

\int \frac{1}{( x + 3 ) ( x + 3 ) ^{2} - 36} d x

\frac{1}{6} arcsec (\frac{1}{6} (x + 3)) + C

Solution steps

\int \frac{1}{( x + 3 ) ( x + 3 ) ^{2} - 36} d x

Apply u-substitution

: \int \frac{1}{u u ^{2} - 36} d u

= \int \frac{1}{u u ^{2} - 36} d u

Apply Trigonometric Substitution

: \int \frac{1}{6} d v

= \int \frac{1}{6} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{6} \cdot 1 \cdot v

Substitute back

= \frac{1}{6} \cdot 1 \cdot arcsec (\frac{1}{6} (x + 3))

Simplify

= \frac{1}{6} arcsec (\frac{1}{6} (x + 3))

Add a constant to the solution

= \frac{1}{6} arcsec (\frac{1}{6} (x + 3)) + C

\int \frac{1 + cos ( 2 x )}{1 - cos ( 2 x )} d x

\int (\frac{2}{y} + y^{2}) e^{x} d x

\int (8 x + 4)^{\frac{3}{2}} d x

\int cos (2 x) sin^{2} (2 x) d x

2^{x^{2}}

What is the integral of 1/((x+3)sqrt((x+3)^2-36)) ?
The integral of 1/((x+3)sqrt((x+3)^2-36)) is 1/6 arcsec(1/6 (x+3))+C